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  1. The problem statement, all variables and given/known data Hi

I am looking at the proof attached for the theorem attached that:

If $s \in R$, then $\sum'_{w\in\Omega} |w|^-s $ converges iff $s > 2$ where $\Omega \in C$ is a lattice with basis ${w_1,w_2}$.

For any integer $r \geq 0 $ :

$\Omega_r := {mw_1+nw_2|m,n \in Z, max {|m|,|n|}=r} $

$\Pi_r := {mw_1+nw_2|m,n \in Z, max {|m|,|n|}=r} $

so that $\Omega = {0} \Cup \Omega_1 \Cup \Omega_2 \Cup....$

Each $\Omega_r$ has cardinality $8r$

QUESTIONS - To prove via the comparison test, we only need to bound from above by a series that converges, so why have we bound from above and below - this is my main question really, why have we bound from above and below

  • Does this proove via both the convergence test and the Weierstass-M test? Since each term in the sequence $|w|^{-s}$ is bound above by a real constant.

  • The definition of the W-M test is $u_n$ a seqence of functions, if for each a $n \in N$ there exists $M_n \in R$ satisfying $|u_n(z)|\leq M_n $ got all $z \in E$ where $u_n : E \to C$ and $\sum M_n$ converges. Here the '$u_n$' are $|w|$ are already taken the absolute value, does this change anything here or the W-M test or does it still apply in the same way?

Many thanks in advance.

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    $|nw_1+m w_2|^{-s}-\int_0^1 \int_0^1 |(n+x)w_1+(m+y)w_2|^{-s}dx dy = O(\ldots)$ – 2017-02-19
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    thanks for your reply, I wasn't looking for another method to it, but to address the questions in the problem, thanks – 2017-02-25

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